Abstract Algebra Midterm 2 F19PL
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Instructions
Start each problem on a new page. Parts of the same problem can be written on
the same page.
The following three problems are independent and each part is worth 4 marks. You
must explain your reasoning and justify carefully everything that you write in
order to get marks. If asked to give an example of a mathematical object satisfying
certain properties, you must explain why your answer satisfies the required properties.
If you want to use a result or theorem from the lecture notes, you must justify that
its assumptions are satisfied. Any statement or result that does not appear in the
lecture notes must be fully justified in order to get marks.
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Problem 1. We consider the set G of 3× 3 matrices with coefficients in Z2 defined as follows:
G :=
{1 a b0 1 c
0 0 1
 | a, b, c ∈ Z2.}
(a) Show that G is a subgroup of GL3(Z2). What is its order?
(b) Show that the following two elements A,B ∈ G do not commute:
A :=
1 1 00 1 0
0 0 1
 , B :=
1 0 00 1 1
0 0 1
 .
Is G a cyclic group?
(c) Show that G is generated by A and B.
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Abstract Algebra Midterm 2 F19PL
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Problem 2. Read the following mathematical text and answer the associated comprehension
questions at the end.
The goal of this text is to prove the following theorem:
Theorem 1. For n > 3, the multiplicative group Z×2n is not cyclic.
We start by proving several intermediate results:
Proposition 2. For m > 1, the additive group Zm contains at most one element
of order 2.
Proof. The proposition is true if m is odd. Indeed, a group of odd order does
not contain any element of order 2 by Lagrange’s Theorem. Thus, it is enough to
consider the case where the order of Zm is of the form m = 2` for some integer
` > 1.
Let a ∈ Z2` be an element of order 2 with a an integer such that 0 6 a 6 2`− 1.
We thus have a + a = 0 in Z2`, i.e. 2a is divisible by 2`. This implies that a is
divisible by `. Since 0 6 a 6 2`− 1, it follows that we either have a = 0 or a = `.
Thus, we either have a = 0, which is the identity element and has order 1, or
a = `, which is an element of order 2. Thus, the only element of order 2 of Z2`
is `.
Proposition 3. Let G be a finite cyclic group. Then G contains at most one
element of order 2.
Proof. Let m be the order of G. We know from the course that a cyclic group
of order m is isomorphic to the additive group Zm, so let us consider a group
isomorphism f : G → Zm. If G does not contain any element of order 2, there
is nothing further to do, so let us assume that G contains an element x of order
2. We have x2 = e in G. We thus have f(x2) = f(e) in Zm, and by properties of
group homomorphisms, this implies that
f(x) + f(x) = 0.
Thus, f(x) is an element of Zm of order at most 2. However, since x is non-trivial,
f(x) cannot have order 1. Thus, it has order exactly 2. By Proposition 2, it thus
follows that f(x) is the unique element of Zm of order 2.
Let y ∈ G be another element of order 2 of G. The same reasoning shows that
f(y) is also the unique element of order 2 of Zm. Thus, f(x) = f(y), and by
injectivity of f , it follows that x = y. Thus, G contains exactly one element of
order 2.
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Abstract Algebra Midterm 2 F19PL
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We are now ready to prove Theorem 1:
Proof of Theorem 1. We know that an element k ∈ Z2n belongs to Z×2n if and only
if k and 2n are coprime, that is, if and only if k is odd. In particular, the elements
a := 2n−1 − 1 and b := 2n−1 + 1
belong to Z×2n , and we have
a2 = (2n−1 − 1)2 = 2n︸︷︷︸
=0
×2n−2 − 2n︸︷︷︸
=0
+ 1 = 1,
and
b2 = (2n−1 + 1)2 = 2n︸︷︷︸
=0
×2n−2 + 2n︸︷︷︸
=0
+ 1 = 1.
Thus, Z×2n contains at least two distinct elements of order 2. Since cyclic groups
contain at most one element of order 2 by Proposition 2, this implies that Z×2n is
not cyclic.
Comprehension questions:
(a) Following the same approach as in the proof of Proposition 2, show that
for an integer m > 1, the additive group Zm either contains no element of
order 3 or contains exactly two elements of order 3.
(b) Explain how Proposition 3 can be used to show that the multiplicative
group Z×99 is not cyclic.
(c) In the proof of Proposition 3, the following statement is not justified:
“However, since x is non-trivial, f(x) cannot have order 1.”
Properly justify this statement.
(d) Show that Theorem 1 does not hold for n = 1 and n = 2. That is, show that
the multiplicative groups Z×2 and Z×4 are cyclic.
(e) The proof of Theorem 1 is thus currently incomplete, as it does not explicitly
use the hypothesis that n > 3. In what parts of the proof is that assumption
implicitly used?
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Abstract Algebra Midterm 2 F19PL
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Problem 3. The goal of this problem is to study the automorphisms of S5, that is, the iso-
morphisms from S5 to itself. Let f : S5 → S5 be an automorphism. To lighten
the notations, we denote by τ the transposition (1 2) ∈ S5.
(a) Show that the permutation f(τ) ∈ S5 has order 2.
(b) Use part (a) to show that f(τ) is either a transposition or a product of two
disjoint transpositions.
(c) Show that a permutation σ ∈ S5 commutes with τ if and only if the permu-
tation f(σ) commutes with f(τ). Deduce that the number of permutations
in S5 that commute with τ and the the number of permutations in S5 that
commute with f(τ) are equal.
Let (a b) ∈ S5 be a transposition, with 1 6 a, b 6 5 distinct. We recall that,
for a permutation σ ∈ S5, the permutation σ ◦ (a b) ◦ σ−1 is also a transposition,
namely
σ ◦ (a b) ◦ σ−1 = (σ(a) σ(b))
(This was proved in Problem Sheet 7, you do not have to prove it here.)
(d) Show that a permutation σ ∈ S5 commutes with the transposition (a b) if
and only we are in one of the following situations:
σ(a) = a and σ(b) = b,
or
σ(a) = b and σ(b) = a.
How many permutations in S5 commute with the transposition (a b)?
(e) Use a similar strategy to determine the number of permutations in S5 that
commute with a product of two disjoint transpositions of the form (a b)(c d)
with 1 6 a, b, c, d 6 5 all distinct.
(f) Use the previous parts to deduce that f(τ) is a transposition.
END OF PAPER
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